find sin(2x), cos(2x), and tan(2x) from the given information. sin(x)= - 8/17, x in quadrant iii sin(2x)=…

Explanation:

Step1: Find cos(x)

Using the identity $\sin^{2}x+\cos^{2}x = 1$, we have $\cos^{2}x=1 - \sin^{2}x$. Given $\sin(x)=-\frac{8}{17}$, then $\cos^{2}x=1-\left(-\frac{8}{17}\right)^{2}=1-\frac{64}{289}=\frac{289 - 64}{289}=\frac{225}{289}$. Since $x$ is in Quadrant III, $\cos(x)<0$, so $\cos(x)=-\frac{15}{17}$.

Step2: Find sin(2x)

Use the double - angle formula $\sin(2x)=2\sin(x)\cos(x)$. Substitute $\sin(x)=-\frac{8}{17}$ and $\cos(x)=-\frac{15}{17}$ into it: $\sin(2x)=2\times\left(-\frac{8}{17}\right)\times\left(-\frac{15}{17}\right)=\frac{240}{289}$.

Step3: Find cos(2x)

Use the double - angle formula $\cos(2x)=\cos^{2}x-\sin^{2}x$. Substitute $\sin(x)=-\frac{8}{17}$ and $\cos(x)=-\frac{15}{17}$: $\cos(2x)=\left(-\frac{15}{17}\right)^{2}-\left(-\frac{8}{17}\right)^{2}=\frac{225}{289}-\frac{64}{289}=\frac{161}{289}$.

Step4: Find tan(2x)

Use the formula $\tan(2x)=\frac{\sin(2x)}{\cos(2x)}$. Substitute $\sin(2x)=\frac{240}{289}$ and $\cos(2x)=\frac{161}{289}$: $\tan(2x)=\frac{\frac{240}{289}}{\frac{161}{289}}=\frac{240}{161}$.

Answer:

$\sin(2x)=\frac{240}{289}$ $\cos(2x)=\frac{161}{289}$ $\tan(2x)=\frac{240}{161}$